> Edit: neighbouring vertices can't be collapsed to a 1-dim string such that each bit x in the string abuts bit y if x is a neighbouring vertex of y (neighbouring = 1 edge away). Works for 1 and 2 dim cubes (line and square) fails for 3. Maybe am missing something.
It sounds like you're missing something. A 3-dimensional cube has 2^3 = 8 vertices:
x y z
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Since there are three dimensions, each vertex has three neighbors. If you imagine the vertex at (0, 1, 1), you can travel across the x-dimension to the neighbor at (1, 1, 1), or across the y-dimension to the neighbor at (0, 0, 1), or across the z-dimension to the neighbor at (0, 1, 0). Obviously, you represent this by flipping the x-, y-, or z-bit in the corresponding bitstring representation. The bitstring representation itself is blindingly straightforward: the vertex at (0, 1, 1) has the representation 011, etc.
It's not really clear to me where you're becoming confused?
> What does 'embedded' even mean, mathematically.
In this case, it means that a complex structure (the cube) includes a simpler structure (the collection of discrete bitstrings). You can analyze the bitstrings by themselves if you want to, or you can imagine them as being the vertices of a cube that may or may not have properties other than its vertices, but however you describe the cube, it will include the bitstrings.